Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
\sqrt {9 - 4\sqrt 5 } - \sqrt {9 + 4\sqrt 5 } \\
= \sqrt {5 - 2.\sqrt 5 .2 + 4} - \sqrt {5 + 2.\sqrt 5 .2 + 4} \\
= \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + 2} \right)}^2}} \\
= \left( {\sqrt 5 - 2} \right) - \left( {\sqrt 5 + 2} \right)\\
= - 4\\
2,\\
\sqrt {6 + 2\sqrt 5 } - \sqrt {29 - 12\sqrt 5 } \\
= \sqrt {5 + 2\sqrt 5 + 1} - \sqrt {20 - 2.2\sqrt 5 .3 + 9} \\
= \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt {{{\left( {2\sqrt 5 - 3} \right)}^2}} \\
= \left( {\sqrt 5 + 1} \right) - \left( {2\sqrt 5 - 3} \right)\\
= 4 - \sqrt 5 \\
3,\\
\sqrt {x - 2\sqrt x + 1} = \sqrt {{{\left( {\sqrt x - 1} \right)}^2}} = \left| {\sqrt x - 1} \right| = \sqrt x - 1\,\,\,\,\,\,\,\,\,\,\,\,\left( {x \ge 1 \Rightarrow \sqrt x \ge 1} \right)\\
4,\\
\sqrt {3x - 2\sqrt {3x} + 1} = \sqrt {{{\left( {\sqrt {3x} - 1} \right)}^2}} = \left| {\sqrt {3x} - 1} \right| = \sqrt {3x} - 1\\
5,\\
2\sqrt {27} - \sqrt {\dfrac{{16}}{9}} - \sqrt {48} - \sqrt {\dfrac{{25}}{9}} \\
= 2.\sqrt {{3^2}.3} - \sqrt {{{\left( {\dfrac{4}{3}} \right)}^2}} - \sqrt {{4^2}.3} - \sqrt {{{\left( {\dfrac{5}{3}} \right)}^2}} \\
= 2.3\sqrt 3 - \dfrac{4}{3} - 4\sqrt 3 - \dfrac{5}{3}\\
= 2\sqrt 3 - 3\\
2,\\
A = \dfrac{{5{x^2} - 2\sqrt 5 .x + 1}}{{\sqrt 5 x - 1}} = \dfrac{{{{\left( {\sqrt 5 .x - 1} \right)}^2}}}{{\sqrt 5 .x - 1}} = \sqrt 5 x - 1\\
x = 25 \Rightarrow A = \sqrt 5 .25 - 1 = 25\sqrt 5 - 1\\
B = \dfrac{{9x - 6\sqrt x + 1}}{{3\sqrt x - 1}} = \dfrac{{{{\left( {3\sqrt x - 1} \right)}^2}}}{{3\sqrt x - 1}} = 3\sqrt x - 1\\
x = 81 \Rightarrow B = 3\sqrt {81} - 1 = 26\\
C = \dfrac{{16 - 40\sqrt x + 25x}}{{4 - 5\sqrt x }} = \dfrac{{{{\left( {4 - 5\sqrt x } \right)}^2}}}{{4 - 5\sqrt x }} = 4 - 5\sqrt x \\
x = 9 \Rightarrow C = 4 - 5\sqrt 9 = - 11
\end{array}\)
